# Star formation

## Homework Statement

I am trying to describe the limit for the minimum radius of a star to first year university level students in order to describe the reason for the size of the Sun. I'm a bit confused by the critical density of the material in the star's core and how we know that the main sequence stars have no degenerate electrons.

## Homework Equations

I think to find the critical density we consider de Broglie wavelength of electrons in terms of lambda = h/mv where m and v are the mass and velocity of the electrons. Any help on how to proceed using de Broglie would be great!

## The Attempt at a Solution

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## Answers and Replies

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Hello! I established a toy model for spinning star in a paper weaks ago, at undergraduate level, below are prat of this work.

Supposing that a star of mass $M$ consists of incompressible fluid of density $\rho$, and it rotates uniformly around its geologically fixed polar axis at a constant angular velocity of $\omega$, relative to an inertial reference $R$ at rest. To fix a 3D Cartesian coordinate system $R_C[O;\vec{i},\vec{j},\vec{k}]$ and a spherical
one $R_S[O;\vec{e_r},\vec{e_\theta},\vec{e_\varphi}]$, who share the same origin which coincides the geometrical center of the star, and the unit vector $\vec{k}$ forms a right-handed spiral mathematical
structure with the spin. Hence, in equilibrium, an arbitrary mass element with coordinates $P(r,\theta,\varphi)$ receives two body forces:

\vec{F_1}=-F_1{\vec{e_r}}

\vec{F_2}=\omega^2r\cos\beta(\cos\beta\vec{e_r}+\sin\beta\vec{e_\theta})

where $\vec{F_1}$ is gravitation from the remaining mass of the star, $\vec{F_2}$ being the centrifugal inertial force, $\beta$
being the declination, $\beta=\frac{\pi}{2}-\theta$. The equation of equilibrium for the very mass element is$[1]$ :

\rho\vec{F}=\nabla p

Considering that

\vec{F}=\vec{F_1}+\vec{F_2}

and the gradient operator in spherical coordinates is:

\nabla=\frac{\partial}{\partial r}\vec{e_r}+\frac{1}{r}\frac{\partial}{\partial\theta}\vec{e_\theta}+\frac{1}{r\sin\theta}\frac{\partial}{\partial\varphi}\vec{e_\varphi}

Then Eq3 turns into
\begin{subequations}

\rho(-F_1+\omega^2r\cos^2\beta)=\frac{\partial p}{\partial r}

\frac{1}{2}\rho\omega^2r\sin2\beta=\frac{1}{r}\frac{\partial
p}{\partial \theta}=-\frac{1}{r}\frac{\partial p}{\partial \beta}

0=\frac{1}{r\sin\theta}\frac{\partial p}{\partial
\varphi}=\frac{\partial p}{\partial \varphi}

\end{subequations}
These are the componential equations of equilibrium under impressible fluid scenario. Immediately, let's get down to the equation of the stellar surface. \\

The surface is a generalized equipotential one, and the potential energy $U_m$ per unit mass is the resultant of its gravitational potential energy $U_1$ and the centrifugal inertial potential energy $U_2$$[2]. U_m=U_1+U_2 Assuming the stellar mass distribution is spherically symmetric, and the gravitational potential energy is set zero for reference in r=+\infty, then \emph{approximately} U_1=-\frac{GM}{r} As to U_2, an alternative approach to \vec{F_2} is: \vec{F_2}=-\nabla U_2=-\frac{\partial U_2}{\partial r}\vec{e_r}-\frac{1}{r}\frac{\partial U_2}{\partial \theta} \vec {e_\theta} Comparison of the coefficients of each orthogonal components in Eq2 and Eq9 lead to \begin{subequations} \frac{\partial U_2}{\partial r}=-\omega^2r\cos^2\beta \frac{\partial U_2}{\partial \theta}=-\frac{1}{2}\omega^2r^2\sin2\beta \text{For \frac{\partial U_2}{\partial \theta}=-\frac{\partial U_2}{\partial \beta}, Eq10b also reads} \frac{\partial U_2}{\partial \beta}=\frac{1}{2}\omega^2r^2\sin2\beta \end{subequations} Hence, the proper differential of U_2 is \begin{split} dU_2&=\frac{\partial U_2}{\partial r}dr+\frac{\partial U_2}{\partial \beta}d\beta\\ &=-\omega^2r\cos^2\beta dr+\frac{1}{2}\omega^2r^2\sin2\beta d\beta\\ &=-d\left(\frac{1}{2}\omega^2r^2\cos^2\beta \right) \end{split} Set the centrifugal potential energy to be zero for reference at r=0 (Noting that \beta=\frac{\pi}{2} corresponds to Riemannian singularity, and lacks generality), after integral we have U_2=-\frac{1}{2}\omega^2r^2\cos^2\beta U_m=U_1+U_2=-\frac{GM}{r}-\frac{1}{2}\omega^2r^2\cos^2\beta As to Eq13, considering that the zero point for U_m, and U_1 and U_2 are selected individually, the resultant U_m is divergent for both r=\infty and r=0. However, because Eq13 is employed for investigations simply at r=\bar{R} (\bar{R} refers to the mean value of the stellar's radii) where U_m is conservative, there's no need to re-asset the zero potential point for U_m here. Since the surface is equipotential, U_m = \text{constant}=C, then -\frac{GM}{r}-\frac{1}{2}\omega^2r^2\cos^2\beta=C Denote R_p to be the radius between the center and one pole( R_p is observable), then insert the boundary condition \{\beta=\frac{\pi}{2},r=R_p\} into Eq14, we have C=-\frac{GM}{R_p} From Eq14 and Eq15, the equation of the surface reads f(r,\beta)=\frac{1}{2}\omega^2r^2\cos^2\beta+\frac{GM}{r}-\frac{GM}{R_p}=0 or f(r,\theta)=\frac{1}{2}\omega^2r^2\sin^2\theta+\frac{GM}{r}-\frac{GM}{R_p}=0 The physical background for the geometrical equations of the surface is the equipotential conditions. According to the symmetric properties, Eq16 or Eq17 refers to the surface of a rotating ellipsoid geometrically, which is generated by the rotation around the minor axis of the ellipse whose major axis is the diameter of the stellar's equator and minor axis is the diameter between two opposite poles. In particular, for the equator, \theta=\frac{\pi}{2}(\beta=0), r=R_E, then from Eq17 we have \frac{R_E-R_P}{R_P}=\eta=\frac{\omega^2R_E^3}{2GM} where \eta is traditionally defined[3] as the flattening of the a rotating ellipsoid. If \eta is slight, the rotating ellipsoid is also called analogous sphere. Eq18 provides a direct perception of the effect of the horizontal expansion due to spin. The faster the stellar's spin and the larger the volume(R_E^3), the more serious the deformation. As for the earth we live on, R_P=6356.766km, R_E=6378.160km, \omega=\frac{2\pi}{24\times 3600}rad\cdot s^{-1}\approx 7.27\times10^{-5} rad\cdot s^{-1}, G=6.67\times10^{-11}N\cdot m^2\cdot kg^{-2}, M=5.796\times 10^{24}kg$$[3]$, hence

\eta_e=\frac{\omega^2R_E^3}{2GM}\approx 0.1773\times10^{-2}

which is about half of the popular value, thus Eq18 is reasonable to a considerable degree. And the flattening of the earth is small, so it can be treated as a analogous sphere in certain astrophysical problems for accuracy.

You can contact tianwj1@gmail.com for a texified pdf file with visual formulae.

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Since you refered to degenerate electrons, then, I don't think there's an ideal way, which is rigorous, yet acceptable for first year university level students.